Question

Let A and B be two nonsingular square matrices of same order such that B is mot an identity matrix and AB^2=BA.If A^3=B^-1A^3B^n then value of n is

Answer 1

B^-1 here means the inverse of B

Answer 2

@ganeshie8

Answer 3

\[A ^{3}=B ^{-1}*A ^{3}*B ^{n}\] is what you have written yes ?

Answer 4

Yeah!

Answer 5

if AB^2=BA, then A^3=B^-1A^3B^n which is n here equals 1

Answer 6

Where did u use the first relation
Ab^2=BA?

Answer 7

Options are 1)4
2).2
3).8
4).9

Answer 8

@kainui

Answer 9

I left multiplied by the inverse of B to obtain:
\[B^{-1}AB^2 = A\]
then I cubed this. Then I used \(AB^2=BA\) to squeeze the A's together since this basically lets you commute A with B. Play around with it, give it your best shot. @knov already gave you a great answer.

Answer 10

Bt we are not having an option as such for n=1..

Answer 11

I m getting n=2..
And it's not right ..why so ??

Answer 12

@kainui i even got n=4 ..
Are there many values of n for which it can be true??

Answer 13

I don't think so since A and B are nonsingular. Can you show your work for how you got it to get to two different things? Maybe there's a mistake in your work I can help you find.

Answer 14

Okay !! And yes it was a mistake on my part ..
I just got n=2..

Answer 15

Shall i show how i got this ??

Answer 16

@kainui

Answer 17

Yeah sure if you like I will happily check your work :D

Answer 18

A^3=B^-1A^3B^n
Multiplying by B
I get,
BA^3=A^3B^n
(BA)A^2=A^3B^n
AB^2A^2=A^3B^n
A^3B^2=A^3B^n

Answer 19

@kainui

Answer 20

I followed all your steps but I don't see how you did the very last step:

AB^2A^2=A^3B^n
A^3B^2=A^3B^n

Are you saying that \(A^2B^2=B^2A^2\) ?

Answer 21

No i applied associativity ..like
(AB)C=A(BC)

Answer 22

Okk ..we don't have to do like that since what u told is definitely right..
Bt how to solve next??

Answer 23

Since BA=AB^2 you can slowly push these Bs through A one at a time:

AB^2A^2=A^3B^n
Just rewriting to be clearer before I substitute:
AB(BA)A=A^3B^n
Now substituting:
ABAB^2A=A^3B^n
now we do the same with this here:
A(BA)B^2A=A^3B^n
AAB^2B^2A=A^3B^n
A^2B^4A=A^3B^n

So now we have 2 of the As over on the left, now you just gotta push that farthest right A through the 4 Bs to the left of it by this same process.

Answer 24

Okay ! Pls wait for a sec or so!! I m checking..

Answer 25

I got this till here!! Now i m making something more substitution further...

Answer 26

I got the ans 8

Answer 27

Thanks @kainui

Answer 28

Yeah that sounds right to me :D